What is the integral of 1/(x^2sqrt(9+x^2)) ?

The integral of 1/(x^2sqrt(9+x^2)) is -(sqrt(9+x^2))/(9x)+C

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integral of 1/(x^2sqrt(9+x^2))

\int \frac{1}{x ^{2} 9 + x ^{2}} d x

- \frac{9 + x ^{2}}{9 x} + C

Solution steps

\int \frac{1}{x ^{2} 9 + x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{9 tan ^{2} ( u )} d u

= \int \frac{sec ( u )}{9 tan ^{2} ( u )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{9} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{9} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= \frac{1}{9} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{9} (- \frac{1}{v})

Substitute back

= \frac{1}{9} (- \frac{1}{sin ( arctan ( \frac{1}{3} x ) )})

Simplify

\frac{1}{9} (- \frac{1}{sin ( arctan ( \frac{1}{3} x ) )}) : - \frac{9 + x ^{2}}{9 x}

= - \frac{9 + x ^{2}}{9 x}

Add a constant to the solution

= - \frac{9 + x ^{2}}{9 x} + C

(x + 2) sin (y) d x + x cos (y) d y = 0

\frac{d y}{d x} = e^{- x^{2}}, y (3) = 5

\int (4 - 3 x)^{5} + e^{2 x + 5} + sin (9 x) d x

y^{^{'}} + \frac{y}{x} = x

\int_{1}^{\infty} \frac{3}{x ^{5}} d x

What is the integral of 1/(x^2sqrt(9+x^2)) ?
The integral of 1/(x^2sqrt(9+x^2)) is -(sqrt(9+x^2))/(9x)+C